Evaluate $\int^{\pi/3}_{0}6\tan x\sec^3x\,dx\,$.
Answer: First note that the derivative of $~\sec x~-$ namely, $~\sec x\cdot\tan x~-$ appears in the integrand. We regroup the factors of the integrand. $ \int^{\pi/3}_{0}6 \tan{x}\cdot \sec^{3}x \ dx = \int^{\pi/3}_{0}6 (\tan{x} \cdot \sec{x} )\cdot \sec^{2}x \ dx$ Now use a $~u$ -substitution with $ u=\sec x~~~~~$ and $~~~~~du =\tan x\cdot\sec x\,dx\,$. It is also appropriate to change the limits of integration. $ x=0~~~~~~~\Rightarrow~~~~~u=\sec\,0=1$ $ x=\dfrac\pi3~~~~~\Rightarrow~~~~~u=\sec\,\dfrac\pi3=2$ $ \int^{\pi/3}_{0}6 (\tan{x} \cdot \sec{x} )\cdot \sec^{2}x \ dx= 6\int_1^2 u^{2} \ du$ This definite integral is straightforward. $ 6\int_1^2 u^{2} \ du=2u^3\Big|_1^2=16-2=14$